Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: AndroidNeox on 09/12/2016 01:41:24

Here be my think’n
Spacetime dilation is a function of total field tension in any region, right?
Thought experiment: Two identical, ideal springs, (A) & (B), each of rest mass M kg. (A) held under tension, storing X Joules of potential energy, (B) under compression, storing X J of P.E.
In this static state, the stored P.E. will add to the rest mass following E = Mc^2, each spring has an inertial mass of N, where N = M + Z, where Z = X/c^2
Agree, so far?
But, they would have different gravitational masses:
Mass of (A): M  Z
Mass of (B): M + Z
Set to oscillating, the springs would radiate energy via spacetime oscillations due to the tension/compression cycle, independent of any motion of masses.
I know people say that a gravity well just dilates time and curves space, but I disagree. If that were the case, c wouldn’t retain the same local value, everywhere. The apparent slowing of a light beam in a gravity well is because the path through space has been lengthened:
(1) Clocks run slower in a gravity well in proportion to the difference in gravitational potential energy, hence time is dilated.
(2) The time required for a light beam to traverse a region in a gravity well is increased in proportion to the difference in gravitational potential energy, hence space is dilated.
I see no reason to think the Shapiro Delay is constrained to low field conditions… it’s a direct consequence of the requirements of Relativity.
A clock within (A) would run faster than a clock within (B). A radioisotope in a diamond anvil press might have its decay detectably slowed by great enough pressure.
Or, am I totally thinking about this wrong?

You are suggesting that a spring under tension weighs more than one that is not. I suppose that's true in theory, but it would be diabolically difficult to prove by experiment because the energy stored in a practical spring (kx^{2}/2) is trivial compared to the energy stored in its mass (mc^{2}). In any case, the spring constant (k) has nothing to do with mass density and the energy required to squash or stretch the spring does not depend the direction of the applied force. An ideal spring that is compressed weighs the same as one that is stretched by the same amount.

You are suggesting that a spring under tension weighs more than one that is not. I suppose that's true in theory, but it would be diabolically difficult to prove by experiment because the energy stored in a practical spring (kx^{2}/2) is trivial compared to the energy stored in its mass (mc^{2}). In any case, the spring constant (k) has nothing to do with mass density and the energy required to squash or stretch the spring does not depend the direction of the applied force. An ideal spring that is compressed weighs the same as one that is stretched by the same amount.
The spring constant does in fact determine the mass density in part. However the total mass of the compressed spring is strictly a function of the total energy of the compressed spring and the proper mass of the spring.

You're picking at straws. Mass density may be a contributing factor in the determination of the spring constant for a practical material, but it's mostly a function of the shape of the atomic structures and the way in which they mesh. In any case, an ideal spring is massless. The point is that the mass of system A (tension) is the same as the mass of system B (compression.)

You're picking at straws. Mass density may be a contributing factor in the determination of the spring constant for a practical material, but it's mostly a function of the shape of the atomic structures and the way in which they mesh. In any case, an ideal spring is massless.
I'm not picking at straws by any means whatsoever. First of all your comment on an ideal massless spring is out of place in this thread because AndroidNeox stated that that spring does have mass by the definition of his stated problem. Second, the subject of this thread is Question about General Relativity, Field Tension, & Gravity. This means that its not a question asked in terms of nonrelativistic physics. That means that we acknowledge that mass of a compressed spring is a function of the stress in the spring. If the spring is massless and we're only talking about the spring and nothing else then its a meaningless question because such a thing does not exist. Third, stress when considering mass is critically important. In fact if we ignore it when trying to calculate the mass of a classical electron (which is not a particle by has a finite size) then we run into a paradox.
In this forum we do not hide the truth. The OP didn't say anything about the energy in the spring or the stress in the spring. Therefore your comment is totally unjustified since with great enough energy and stress the mass increase is significant. Otherwise your comment is equivalent to saying that we cannot detect gravitational time dilation between two points separated by inches, which is no longer true.
This becomes very important in general relativity. Especially when one considers that stress could be the cause of the accelerated expansion of the universe.
AndroidNeox  If the spring is being compressed by an external force then its mass depends on the stress. If its being held together by, say, a clasp then the spring/clasp system as a whole is not a function of stress because the two stresses in the spring and clasp cancel out.
Mike  In the future please don't try to say that someone is picking at straws merely because you think that a contribution to something is small. Especially when talking about relativity. It brings bad karma to the forum. :)

It does actually make sense to talk about the mass of a massless spring. You just can't get it to oscillate unless you attach some mass to it. You are correct to say that the spring constant can be arbitrarily large and the masses can be arbitrarily small, but it makes no difference. The point is that the energy required to compress an ideal spring is the same as that required to stretch it by the same amount.

It does actually make sense to talk about the mass of a massless spring. You just can't get it to oscillate unless you attach some mass to it.
You are still incorrect. First of all a spring can oscillate with no mass attached. Where did you get the notion that it can't? If the spring has mass then it most certainly can. In fact this is a problem found in many classical mechanics texts.
Also note that I said massless spring and not mass on a massless spring. Do you understand the difference? The question posed here is about a spring of mass N which is in general not zero.
You are correct to say that the spring constant can be arbitrarily large and the masses can be arbitrarily small, but it makes no difference. The point is that the energy required to compress an ideal spring is the same as that required to stretch it by the same amount.
Clearly your understanding of relativity is limited. A full consideration of relativistic mechanics requires what is known as the stressenergymomentum
(SEM) tensor. And to calculate mass density requires the use of the SEM tensor. If you're not familiar with this object then see: http://www.newenglandphysics.org/physics_world/sr/energy_momentum_tensor.htm
A good GR text will give problems in which you have to use the stress to determine the mass of an object. Mould's text Basic Relativity is a good example. Before you post more incorrect claims based on your lack of understanding of the subject I suggest that you look at his text and learn the subject matter in more depth. Especially before you claim that I'm wrong.

Here be my think’n
Spacetime dilation is a function of total field tension in any region, right?
Thought experiment: Two identical, ideal springs, (A) & (B), each of rest mass M kg. (A) held under tension, storing X Joules of potential energy, (B) under compression, storing X J of P.E.
In this static state, the stored P.E. will add to the rest mass following E = Mc^2, each spring has an inertial mass of N, where N = M + Z, where Z = X/c^2
Agree, so far?
But, they would have different gravitational masses:
Mass of (A): M  Z
Mass of (B): M + Z
Set to oscillating, the springs would radiate energy via spacetime oscillations due to the tension/compression cycle, independent of any motion of masses.
I know people say that a gravity well just dilates time and curves space, but I disagree. If that were the case, c wouldn’t retain the same local value, everywhere. The apparent slowing of a light beam in a gravity well is because the path through space has been lengthened:
(1) Clocks run slower in a gravity well in proportion to the difference in gravitational potential energy, hence time is dilated.
(2) The time required for a light beam to traverse a region in a gravity well is increased in proportion to the difference in gravitational potential energy, hence space is dilated.
I see no reason to think the Shapiro Delay is constrained to low field conditions… it’s a direct consequence of the requirements of Relativity.
A clock within (A) would run faster than a clock within (B). A radioisotope in a diamond anvil press might have its decay detectably slowed by great enough pressure.
Or, am I totally thinking about this wrong?
Here be my think’n
Spacetime dilation is a function of total field tension in any region, right?
Thought experiment: Two identical, ideal springs, (A) & (B), each of rest mass M kg. (A) held under tension, storing X Joules of potential energy, (B) under compression, storing X J of P.E.
In this static state, the stored P.E. will add to the rest mass following E = Mc^2, each spring has an inertial mass of N, where N = M + Z, where Z = X/c^2
Agree, so far?
But, they would have different gravitational masses:
Mass of (A): M  Z
Mass of (B): M + Z
Set to oscillating, the springs would radiate energy via spacetime oscillations due to the tension/compression cycle, independent of any motion of masses.
I know people say that a gravity well just dilates time and curves space, but I disagree. If that were the case, c wouldn’t retain the same local value, everywhere. The apparent slowing of a light beam in a gravity well is because the path through space has been lengthened:
(1) Clocks run slower in a gravity well in proportion to the difference in gravitational potential energy, hence time is dilated.
(2) The time required for a light beam to traverse a region in a gravity well is increased in proportion to the difference in gravitational potential energy, hence space is dilated.
I see no reason to think the Shapiro Delay is constrained to low field conditions… it’s a direct consequence of the requirements of Relativity.
A clock within (A) would run faster than a clock within (B). A radioisotope in a diamond anvil press might have its decay detectably slowed by great enough pressure.
Or, am I totally thinking about this wrong?
I sure missed this post when it was made. What a gem.

What needs to be determined is the Shapiro delay of gravitational waves. Once this is determined then some theories can be discounted.

To cut to the chase on the spring idea.
http://www.dummies.com/education/science/physics/howtocalculateaspringconstantusinghookeslaw/
We have a value in Newtons per metre for k. This depends upon the mass which is unimportant for objects in free fall. Objects accelerate at the same rate independent of mass. The value of the jerk of the gravitational field becomes important and can be related to the restoring force. Energy is lost and the spring eventually stops oscillating. How can we make a connection between this and the tidal forces of the field?

Spacetime dilation is a function of total field tension in any region, right?
There’s no such thing as spacetime dilation. I think you’re confusing this with gravitational time dilation which relates the rate of clocks at different gravitational potentials.
Thought experiment: Two identical, ideal springs, (A) & (B), each of rest mass M kg. (A) held under tension, storing X Joules of potential energy, (B) under compression, storing X J of P.E.
In this static state, the stored P.E. will add to the rest mass following E = Mc^2, each spring has an inertial mass of N, where N = M + Z, where Z = X/c^2
An ideal spring is defined as a spring which has no mass.
Agree, so far?
No. :)
But, they would have different gravitational masses:
Mass of (A): M  Z
Mass of (B): M + Z
No. That is incorrect for the reasons stated above.
I know people say that a gravity well just dilates time and curves space, but I disagree.
While it’s true that gravity can cause the effect of gravitational time dilation there need not be a curvature in spacetime associated with it. If there is then gravity is not the cause of it. Curved spacetime is nothing more or less than tidal gradients in the gravitational field. One doesn’t say that gravity cause tidal gradients but rather that a particular kind of variation in gravity is a tidal gradient.

AndroidNeox seems to have lost interest, but I think he was pondering 3 points:
1) The mass equivalence of spring tension.
2) The energy in a gravitational wave.
3) Variable light speed in a gravitational field.
PmbPhy raised one more:
4) Oscillation of an ideal (i.e. massless) spring.
The 1st harkens back to Einstein's thought experiment in which a photon is trapped inside a container with a perfectly reflective inner surface and eventually escapes through a small hole. The box loses weight when the photon escapes so it contributes to the mass of the system even though it is massless. The same is true of a massless spring and the upshot is that spring tension is tantamount to mass.
To PmbPhy's point, a massless spring cannot oscillate because it has no kinetic energy when its parts are in motion. You need some mass to carry the momentum. You can attach masses to the ends of the spring or distribute them along it's length. It makes no never mind to the end result, but the distributed case is harder to analyze.
Point #3 is surprising to some people, but it is an undisputed consequence of the Schwarzschild solution.
It is point #2 that I find intriguing because, unlike a planet in orbit around its sun, the system loses energy to gravitational waves.

The 1st harkens back to Einstein's thought experiment in which a photon is trapped inside a container with a perfectly reflective inner surface and eventually escapes through a small hole. The box loses weight when the photon escapes so it contributes to the mass of the system even though it is massless. The same is true of a massless spring and the upshot is that spring tension is tantamount to mass.
Not quite. I explained what the SEM tensor is in the URL I gave above. In Einstein's box thought experiment the stress does not contribute to mass. Please understand that its really stress that contributes to mass. It merely appears that you're not familiar with this all too true fact. Most physics enthusiasts aren't because its not something taught in most relativity texts but only in advanced relativity texts, and even then not in all.
If you don't read Mould's text you may never come to understand this fact.
Tell me something. Are you aware of the fact that the active gravitational mass density of a source is a function not only of energy but of pressure/stress as well? Look it up.

For the uninitiated, SEM = StressEnergyMomentum. It is certainly true that GR represents mass as a pressure gradient, but that's beside the point. Relativity has nothing to say about the mass equivalence of spring tension or the weight of a box that contains a photon.

It is certainly true that GR represents mass as a pressure gradient, but that's beside the point. Relativity has nothing to say about the mass equivalence of spring tension or the weight of a box that contains a photon.
Mike  Do you understand that merely claiming that you're right when someone has given you a resource to show you the correct physics is not a scientifically valid line of reasoning?
All of those claims are wrong. Relativity does have something to say about the contribution of stress to mass. As I've explained, numerous times now, you simply don't know what you're talking about.
In fact I derive an expression for it in my paper on mass which is online at
https://arxiv.org/abs/0709.0687
Go to page 8 and see derivation, equations 6, 7 and 8. See also
Relativity, Thermodynamics and Cosmology, Richard C. Tolman, Dover Pub, (1987), p. 68, Eq. (35.9)
which can be downloaded from: http://bok.org/book/509061/d3d58b
As I keep trying to explain to Mike, this is a well known fact in relativity and can be found in a good relativity text such as that my Mould, Tolman etc.
Mike: Please stop posting these invalid claims of yours and learn the correct physics. Its not a simple as you thought it was.

I may get the Mould text. I hadn't heard of it before.

Calm down PmbPhy. All I'm saying is that you can't derive Hooke's law from relativity. The spring constant is a materials property. It may involve spacetime tension at the microscopic level, but it's not that simple because gravity is an insignificant factor in atomic physics. The point of the ideal spring in AndroidNeox's thought experiment is to get the masses oscillating so they generate gravitational waves. The spring works against gravity. It is not itself gravity.

But, they would have different gravitational masses:
Mass of (A): M  Z
Mass of (B): M + Z
No. It doesn't matter whether you stretch or compress a spring: the added energy is positive. Energy is a scalar, not a vector!

You are suggesting that a spring under tension weighs more than one that is not. I suppose that's true in theory, but it would be diabolically difficult to prove by experiment because the energy stored in a practical spring (kx^{2}/2) is trivial compared to the energy stored in its mass (mc^{2}). In any case, the spring constant (k) has nothing to do with mass density and the energy required to squash or stretch the spring does not depend the direction of the applied force. An ideal spring that is compressed weighs the same as one that is stretched by the same amount.
The spring constant does in fact determine the mass density in part. However the total mass of the compressed spring is strictly a function of the total energy of the compressed spring and the proper mass of the spring.
I just realized that I read the 1st sentence of this reply backwards. I thought you were saying that mass density determines the spring constant rather than the other way around. My bad. Alan's comment (above) sums up what I was trying to say in my 1st comment, but correcting that error (scalar vs. vector) doesn't answer the 2nd part of AndroidNeox's question, which has to do with conservation of energy when one of the massspringmass systems is oscillating. If the spring is frictionless, does the system lose energy to gravitational waves?

But, they would have different gravitational masses:
Mass of (A): M  Z
Mass of (B): M + Z
No. It doesn't matter whether you stretch or compress a spring: the added energy is positive. Energy is a scalar, not a vector!
Energy is not a scalar since, by definition as used in relativity etc., a scalar is a tensor of rank zero, i.e. an invariant.
Note: It's unfortunate that the term scalar is an overloaded term. E.g. in chemistry and high school physics texts, scalar is defined as a number but in all advanced physics texts its defined as a tensor of rank zero. In Lanczo's text on classical mechanics a "number" which has different values in different frames its called a covariant quantity whereas quantities such as proper mass are called invariant quantities.

Point taken, but Alan was obviously trying to distinguish vectors from numbers, not variants from invariants. With regards to energy conservation, Feynman's sticky bead argument (https://en.wikipedia.org/wiki/Sticky_bead_argument) seems to be the consensus view. The system does indeed lose energy to gravitational waves, which means they can be harnessed to do work.

If a system can lose energy due to gravitational waves and these waves can be harnessed to do work, does the reciprocal of this mean that a gravitational wave on the scale of the LIGO event will add energy to a system (earth) that it hits (for the duration of the hit)?

That's the idea. Unlike solar cells, it would work in the dark. But we're talking about very low power levels unless you get up close and personal with the source. Not nearly enough to heat your house or drive your car here on Earth. Not a practical means of communication either because you can't focus the beam or modulate the source signal. It might make a good accident avoidance system for intergalactic travel. I read somewhere that gravitational waves can travel between universes, too. I can't imagine a practical use for that though.

Point taken, but Alan was obviously trying to distinguish vectors from numbers, not variants from invariants.
That was understood. That's why what I posted was a Note. Its intended to be in addition to something. In this case its to warn that when one comes across the term scalar that one understands what it really means in advanced physics and mathematics.

Point taken, but Alan was obviously trying to distinguish vectors from numbers, not variants from invariants.
That was understood. That's why what I posted was a Note. Its intended to be in addition to something. In this case its to warn that when one comes across the term scalar that one understands what it really means in advanced physics and mathematics.
Agreed. Unambiguous terminology is half the battle. I experienced a prime example of that recently when I was schooled by a formally trained expert about the definition of the Schwarzschild event horizon. I thought it was a distance measured in Schwarzschild coordinates, but he said it is in fact the future of the causal past of future null infinity. How's that for terminology? Here's a good reference if you want to decode that definition: https://arxiv.org/pdf/0811.0354v1.pdf